So the path integral tells us to integrate over all

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Chapter 10 / Exercise 38
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is over all space-time. So the path integral tells us to integrate over all classical field configurations Φ . Note that Φ does not just consist of the one-particle states, it can have 2-particle states, etc. We can remember this by drawing pictures for the paths – including disconnected bubbles – as we would using Feynman rules. Actually, we really sum over all kinds of discontinuous, disconnected random fluctuations. In perturbation theory, only paths corresponding to sums of states of fixed particle number contribute. Non-perturbatively, for example with bound states or situations where multiple soft photons are relevant, particle number may not be a useful concept. The path integral allows us to perform calculations in non-perturbative regimes. 2.4 Classical limit As a first check on the path integral, we can take the classical limit. To do that, we need to put back planckover2pi1 which can be done by dimensional analysis. Since planckover2pi1 has dimensions of action, it appears as ( 0; t f | 0; t i ) = N integraldisplay D Φ( x,t ) e i planckover2pi1 S [Φ] (34) 2. If V depended on π ˆ and φ ˆ , there might be an ordering ambiguity; this is no different than in the non-rela- tivistic case and it is conventional to define the Hamiltonian to be Weyl-ordered with the π ˆ operators all to the left of the φ ˆ operators. 6 Section 2
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Chapter 10 / Exercise 38
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Using the method of stationary phase we see that in the limit planckover2pi1 0 , this integral is dominated by the value of Φ for which S [Φ] has an extremum. But δS = 0 is precisely the condition which determines the Euler-Lagrange equations which a classical field satisfies. Therefore the only con- figuration which contributes in the classical limit is the classical solution to the equations of motion. In case you are not familiar with method of stationary phase (also known as the method of steepest descent ), it is easy to understand. The quickest way is to start with the same integral without the i : integraldisplay D Φ( x,t ) e 1 planckover2pi1 S [Φ] (35) In this case, the integral would clearly be dominated by the Φ 0 where S [Φ] has a minimum; everything else would give a bigger S [Φ] and be infinitely more suppressed as planckover2pi1 0 . Now, when we put the i back in, the same thing happens, not because the non-minimal terms are zero, but because away from the minimum you have to sum over phases swirling around infinitely fast. When you sum infinitely swirling phases, you also get something which goes to zero when com- pared to something with a constant phase. Another way to see it is to use the more intuitive case with e 1 planckover2pi1 S [Φ] . Since we expect the answer to be well-defined, it should be an analytic func- tion of Φ 0 . So we can take planckover2pi1 0 in the imaginary direction, showing that the integral is still dominated by S 0 ] .

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